# 1998 AHSME Problems/Problem 25

## Problem

A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0)$, and $(7,3)$ is matched with $(m,n)$. Find $m+n$. $\mathrm{(A) \ }6.7 \qquad \mathrm{(B) \ }6.8 \qquad \mathrm{(C) \ }6.9 \qquad \mathrm{(D) \ }7.0 \qquad \mathrm{(E) \ }8.0$

## Solution

### Solution 1

The line of the fold is the perpendicular bisector of the segment that connects $(0,2)$ and $(4,0)$. The point $(m,n)$ is the image of the point $(7,3)$ according to this axis. The situation looks as follows. $[asy] size(200); defaultpen(0.8); pair A=(0,2), B=(4,0), C=(7,3); pair u=(1,2), S=(A+B)*0.5; pair T = intersectionpoint(S -- (S+4*u), C -- (C+4*(A-B)) ); pair D = 2*T-C; draw (A--B); draw ( (S-u) -- (S+4*u), dashed ); draw ( S -- C, black, Arrow ); draw ( S -- D, black, Arrow ); draw ( C -- D, Dotted ); dot(A); dot(B); dot(C); dot(D); label("$$A(0,2)$$",A,SW); label("$$B(4,0)$$",B,S); label("$$S$$",S,WSW*1.3); label("$$T$$",T,ENE*1.3); label("$$C(7,3)$$",C,SE); label("$$D(m,n)$$",D,NE); [/asy]$

Now, we will compute the coordinates of the point $D$, using the following facts:

• The triangles $SCT$ and $SDT$ are congruent.
• $|SD| = |SC|$
• $m$ is positive

As the triangles $SCT$ and $SDT$ are congruent, their areas are equal. The area of the triangle $SCT$ is $1/2$ of the size of the vector product $\overrightarrow{SC}\times\overrightarrow{ST}$, and the area of $SDT$ is $1/2$ of the size of $\overrightarrow{ST}\times\overrightarrow{SD}$.

We get that $\overrightarrow{SC}\times\overrightarrow{ST} = \overrightarrow{ST}\times\overrightarrow{SD}$.

The equality remains valid if we multiply the vector $\overrightarrow{ST}$ by any constant. In other words, instead of $\overrightarrow{ST}$ we can use any vector with the same direction.

The axis of symmetry is perpendicular to $\overrightarrow{AB}$. Thus its direction is $(2,4)=(1,2)$.

We get that $\overrightarrow{SC}\times (1,2) = (1,2) \times\overrightarrow{SD}$.

Substituting the coordinates $\overrightarrow{SC}=(5,2)$ and $\overrightarrow{SD}=(m-2,n-1)$ we get $5\cdot 2 - 2\cdot 1 = 1\cdot (n-1) - 2\cdot(m-2)$. This simplifies to $n=2m+5$.

We just discovered that the coordinates of $D$ are $(m,2m+5)$. We will now use the second two facts mentioned above to find $m$.

We have $|SD| = |SC|$ and therefore $|SD|^2 = |SC|^2$. We know that $|SC|^2 = 5^2 + 2^2 = 29$, and $|SD|^2 = (m-2)^2 + (2m+4)^2$. Simplifying, we get the equation $5m^2 + 12m - 9 = 0$. This has exactly one positive root $m=0.6$.

It follows that $D=(0.6,6.2)$, and that $m+n = 6.2 + 0.6 = \boxed{6.8}$.

### Solution 2

Note that the fold is the perpendicular bisector of $(0,2)$ and $(4,0)$. Thus, the fold goes through the midpoint $(2,1)$.

The fold also has a slope of $-\frac{1}{m}$, where the $m$ is the slope of the line connecting these two points. We find $m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}$. Thus, the slope of the fold is $2$ and goes through $(2,1)$, so the equation of the fold is $y = 2x - 3$.

We want the line connecting $(7,3)$ and $(m,n)$ to have the same fold as a perpendicular bisector. The slope should be $-\frac{1}{2}$, so we get $\frac{n - 3}{m - 7} = -\frac{1}{2}$, which leads to $m = 13 -2n$.

We also want $y = 2x - 3$ to bisect the segment from $(7,3)$ to $(m,n)$. Thus, the midpoint $(x,y) = (\frac{7 + m}{2}, \frac{3 + n}{2})$ must lie on the line. Plugging into the equation of the line, we find $\frac{3 + n}{2} = (7 + m) - 3$, which simplifies to $n = 5 + 2m$

Solving the system of two equations in two variables $m = 13 - 2n$ and $n = 5 + 2m$ gives $m = \frac{3}{5}$ and $n = \frac{31}{5}$ , for a sum of $\boxed{\mathrm{(B) \ }6.8}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 